Changing structures over time 3

[Danielsp]

Hello folks

I am a civil Engineer, M.SC. in structures Engineering and I model and design structures, mostly using FEM.

Right now I am interested in static modeling structures that change over time. Some cases are really simple, for instance:

1) If you load a structure then remove some part of it while keeping the same load, then the new moments, shear forces and strains can be easily obtained by simply ignoring the original structure. You load the new structure and that's it.

2) If you apply Load 1 on Structure 1, then add a structural element to it and finally apply a new Load 2, then the final state of forces and displacements is obtained by adding L1 on Structure 1 to L2 on structure 2.

So far, no big deal.
But what if after the first load you add a new part to the structure while loaded and then remove an old one?

 

[Danielsp]

Another way to do this would be to replace the right support by its force in the original structure, 0,93kN downwards, so it would look like this (spans are 3m wide each).

But demolishing the right span amounts to the same as cutting the right support lose, so I add the necessary load to cancel its effect, which is

When adding both, I arrive at the same results as the other method

 

[Brad805]

Is the sequence below correct? At the point in time you cut the member in stage 1 the end of the remaining section undergo a rotation. When you install the new segments you will build whatever end rotation you have at that time into your joint. I do not see the stress changing as you show unless you deform the new sections somehow during install. Further changes to the boundary conditions or loading will change how it behaves, but the 10kN force always seems present.

 

[Danielsp]

No, Brad. As I pointed out, you demolish after building. If it was the other way around, it would be trivial. And there is no rebuilding.

 

[rb1957]

ok, now some pictures.

but the original 2nd structure has a moment of 3.33 at the LH support,

but your method seems to say 1.04 ?

ie the last 2 pix don't equal the first, no?

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[rb1957]

or are you trying to derive the reactions for the beam on 4 supports ??

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[rb1957]

why would the last pic ...

be different to the first ??

the RH unloaded overhang does nothing to the problem ?

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[Brad805]

Ok, I get it now. If you add the x1/x2/x3 values I would not be surprised if someone would not check this. It is a very easy problem to run if you have a better staged analysis package.

Word of advice. Start with a simple sketch in the future. While you are heavily interested in the problem others come here for different reasons. Not many want to wade thru endless textual descriptions when a decent drawing will convey the question very quickly.

 

[Danielsp]

rb, the 3,33 and 5,56 pic shows what WOULD be the bending moments IF the 10kN load had just added to the two left spans in the usual manner: no history, no previous deformations, no existing displacements or rotations, no initial 3rd span, only a plain 2-span problem. Just for the sake of comparison with the actual values at the end.

 

[driftLimiter]

Let me see if I understand your example please correct me where I go astray.

1) Load structure in original (2) span configuration.
2) Build in new span to the left.
3) Remove the span on the right.

You are assuming that when you install the span on the left, that it has no forces in because it is joined to the already rotated end of the original span.

So theoretically, what would normally be a straight and continuous joint becomes a 'kinked' joint that is still fully fixed. And if you say unload the structure after making that joint you would have stresses and deformations 'locked in'. If you twist the end of the new span to match the rotation of the deflected beam then it would bounce back to neutral after you unload it no locking in.

In any case no matter what the structure has locked in it, external and internal equilibrium has to be satisfied.

 

[Danielsp]

Brad, I didnt use a drawing because I was afraid it would tie the discussion to the specific case.
The real situation I am analyzing, btw, looks very little like the one I showed. It is a slab, a hole of which will be demolished after other regions are reinforced. Actually, after the initial reinforcement there will be 3 demolishing+rebuilding stages. So, I hope all we have discussed really is a general principle.

 

[Danielsp]

The building order and your account of my assumption are correct, drift.

 

[rb1957]

lets label the four supports ... A, B, C, D (for the LH).

so 1) has B, C, D effective. It could have an unloaded overhang to A without any change.

and you're trying to get to 2) (A, B, C loaded) by saying ...
1) starting with 1) ... with PB1, PC1, PD1 and MC1, then
2) apply -PD to 2) (with the RH overhang) and get PA2, PB2, PC2, and MB2 and MC2, then
3) add together ... PA = PA2, PB = PB1+PB2, PC = PC1+PC2, PD = PD-PD = 0, MB = MB2, MC = MC1+MC2 = 0

I see the logic, but I don't think so. It is intriguing that MC2 = -MC1.

I think if you want to go from 1) (BCD) to 2) (ABC) you need to go through 4) (ABCD) and superimpose 2) with -PD.


"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[driftLimiter]

If we could model the joint with a predefined rotation then we could solve your problem quite nicely, not sure if this feature is available....

The new span will have the same rotational stiffness so the final structure will displace proportionally and elastically as it should, but we have already locked a rotation into the joint which would tend to increase deflections in the final condition.

/res.cloudinary.com/engineering-com/image/upload/v1659631029/tips/STAGED-1_ew5fh2_bolxx4.png]

 

[Danielsp]

Drift, that would be easy to model. Just use a fixed instead of pinned support. But it would just be unfeasible to build.

 

[rb1957]

ok, good point ... but what are you building that you need to jump through these hoops ?

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[driftLimiter]

If that is hard to build and easy to model, then I'm lost as to what your thinking is easy to build and what is hard to model about this problem.

You agreed when I said

You are assuming that when you install the span on the left, that it has no forces in because it is joined to the already rotated end of the original span.

If you model that joint as a fixed then this entire problem is moot. I feel as if we are talking around and around here not getting any closer to a solution because the goalposts keep moving.

 

[rb1957]

I wonder if we should be using the 3 moment equation, modelling the beam as interacting spans ?

1) 2 spans, BC and CD; with load P
2) 2 spans, AB and BC; with load P
3) 3 spans, AB, BC, and CD; with load P
4) 2 spans, AB and BC; with load -D3 (the negative load from support D of the three span beam)
3) + 4) = 2) ... maybe !?

I think you're taking the negative of load D from 1) and applying it to ABC, then adding this to 1) to get 2). And I don't think that's right.

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[Danielsp]

rb, It's a 35m diameter 2m thick slab at the bottom of a subway access well 33m deep.
There is already a 40cm slab in place but it's flat and part of final slab will be built below the existing slab and part over it (yeah, the 2m slab has a lower level some 2,5m below and it is connected to the upper level).
We can't simply remove the existing slab because it's actually working as a 2D strut for the concrete walls around it.
So the solution is, build the upper part of the 2m slab on top of the 40cm slab, then open a hole on the thin slab, build the 2m slab region below that hole. Rinse and repeat 3x because it would be too big of a hole to break on our 40cm strut slab.

 

[Danielsp]

drift, I guess steveh49 already solved the problem.
I created an example based on his ideas at 4 Aug 22 11:50.

 

[rb1957]

ok, so now we have the problem ... how to extend a slab. I'll leave that to you Civils, but why not assume two independent slabs (a pin joint) ?

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.